##### Find the x-value(s) of the relative maxima and relative minima

 Calculus Tutor: None Selected Time limit: 1 Day

Jul 15th, 2015

First, we rewrite the 32/x by using this power law:   1/x = 1/x^1 = x^-1

f(x) = 2x^2 + 32/x + 7  ------------> f(x) = 2x^2 + 32x^-1 + 7

Then we find the derivative of the function. Let's remember the derivative of a power:   (x^n)' = n*x^n-1 ; where n is the exponent of the x. By the way, let's remember also that the derivative of any constant is 0.

f'(x) = (2x^2 + 32x^-1 + 7)'  ------> f'(x) = (2x^2)' + (32x^-1)' + (7)' ----> f'(x) = 2(x^2)' + 32(x^-1)' + (7)'

f'(x) = 2(2x^2-1) + 32(-1x^-1-1) + 0  --------> f'(x) = 4x^1 - 32x^-2   --------> f'(x) = 4x - 32x^-2

f'(x)= 4x - 32/x^2

Then we equal the derivative to zero and solve for x.

4x - 32/x^2 = 0  -----------> 4x*x^2 - 32/x^2 * x^2 = 0 ----------->  4x^3 - 32 = 0 ------------> 4x^3 - 32 + 32 = 0 + 32

4x^3 = 32  ----------->  4x^3/4 = 32/4 ------------> x^3 = 8   ------------>  (x^3)^1/3 = (8)^1/3  --------> x = 8^1/3

Or x = cubic root(8) ---------->  x = 2

Then we find the second derivative.

f''(x) = (4x - 32x^-2)' = (4x)' - (32x^-2)' = 4(x)' - 32(x^-2)' = 4(1) - 32(-2)x^-2-1 = 4 + 64x^-3

f''(x) = 4 + 64/x^3

Then we enter x = 2 into the second derivative and if it is greater than zero (positive) then it is a minimum

and if it is less than zero (negative) then it is a maximum.

f''(2) = 4 + 64/2^3 = 4 + 64/8 = 4 + 8 = 12 > 0 (positive). Then x = 2 is a minimum and we don't have any maximum.

Relative maxima: DNE (we don't have it).

Relative minima: x = 2

Please let me know if you have any doubt or question.

Jul 15th, 2015

...
Jul 15th, 2015
...
Jul 15th, 2015
May 28th, 2017
check_circle